Download 8–3 z Test for a Mean

Section 8–3 z Test for a Mean
8–3
Objective
5
Test means for large
samples, using the
z test.
405
z Test for a Mean
In this chapter, two statistical tests will be explained: the z test, used to test for the mean
of a large sample, and the t test, used for the mean of a small sample. This section
explains the z test, and Section 8–4 explains the t test.
Many hypotheses are tested using a statistical test based on the following general
formula:
Test value 冸observed
value 冹 冸expected value冹
standard error
The observed value is the statistic (such as the mean) that is computed from the sample
data. The expected value is the parameter (such as the mean) that one would expect to
obtain if the null hypothesis were true—in other words, the hypothesized value. The
denominator is the standard error of the statistic being tested (in this case, the standard
error of the mean).
The z test is defined formally as follows.
The z test is a statistical test for the mean of a population. It can be used when n 30,
or when the population is normally distributed and s is known.
The formula for the z test is
z
X m
sⲐ兹n
where
X sample mean
m hypothesized population mean
s population standard deviation
n sample size
For the z test, the observed value is the value of the sample mean. The expected value
is the value of the population mean, assuming that the null hypothesis is true. The denominator s兾兹n is the standard error of the mean.
The formula for the z test is the same formula shown in Chapter 6 for the situation
where one is using a distribution of sample means. Recall that the central limit theorem
allows one to use the standard normal distribution to approximate the distribution of sample means when n 30. If s is unknown, s can be used when n 30.
Note: The student’s first encounter with hypothesis testing can be somewhat challenging and confusing, since there are many new concepts being introduced at the same
time. To understand all the concepts, the student must carefully follow each step in the
examples and try each exercise that is assigned. Only after careful study and patience
will these concepts become clear.
As stated in Section 8–2, there are five steps for solving hypothesis-testing problems:
Step 1
State the hypotheses and identify the claim.
Step 2
Find the critical value(s).
Step 3
Compute the test value.
Step 4
Make the decision to reject or not reject the null hypothesis.
Step 5
Summarize the results.
Example 8–3 illustrates these five steps.
8–15
406
Chapter 8 Hypothesis Testing
Speaking of
Statistics
RD HEALTH
This study found that people who used
pedometers reported having increased
energy, mood improvement, and weight
loss. State possible null and alternative
hypotheses for the study. What would be
a likely population? What is the sample
size? Comment on the sample size.
Step to It
I
T FITS in your hand, costs less than $30, and will make
you feel great. Give up? A pedometer. Brenda Rooney,
an epidemiologist at Gundersen Lutheran Medical Center
in LaCrosse, Wis., gave 500 people pedometers and asked
them to take 10,000 steps—about five miles—a day.
(Office workers typically average about 4000 steps a day.)
By the end of eight weeks, 56 percent reported having
more energy, 47 percent improved their mood and 50
percent lost weight. The subjects reported that seeing their
total step-count motivated them to take more.
— JENNIFER BRAUNSCHWEIGER
Source: Reprinted with permission from the April 2002 Reader’s Digest.
Copyright © 2002 by The Reader’s Digest Assn. Inc.
Example 8–3
A researcher reports that the average salary of assistant professors is more than $42,000.
A sample of 30 assistant professors has a mean salary of $43,260. At a 0.05, test the
claim that assistant professors earn more than $42,000 a year. The standard deviation of
the population is $5230.
Solution
Step 1
State the hypotheses and identify the claim.
H0: m $42,000
and
H1: m $42,000 (claim)
Step 2
Find the critical value. Since a 0.05 and the test is a right-tailed test, the
critical value is z 1.65.
Step 3
Compute the test value.
X m $43,260 $42,000
1.32
z
s
兹n
$5230
兹30
Make the decision. Since the test value, 1.32, is less than the critical value,
1.65, and is not in the critical region, the decision is not to reject the null
hypothesis. This test is summarized in Figure 8–13.
Step 4
Figure 8–13
0.45
Summary of the z Test
of Example 8–3
Do not
reject
Reject
0.05
0
Step 5
8–16
1.32 1.65
Summarize the results. There is not enough evidence to support the claim
that assistant professors earn more on average than $42,000 a year.
407
Section 8–3 z Test for a Mean
Comment: Even though in Example 8–3 the sample mean, $43,260, is higher than
the hypothesized population mean of $42,000, it is not significantly higher. Hence, the
difference may be due to chance. When the null hypothesis is not rejected, there is still a
probability of a type II error, i.e., of not rejecting the null hypothesis when it is false.
The probability of a type II error is not easily ascertained. Further explanation about
the type II error is given in Section 8–7. For now, it is only necessary to realize that the
probability of type II error exists when the decision is not to reject the null hypothesis.
Also note that when the null hypothesis is not rejected, it cannot be accepted as true.
There is merely not enough evidence to say that it is false. This guideline may sound a
little confusing, but the situation is analogous to a jury trial. The verdict is either guilty
or not guilty and is based on the evidence presented. If a person is judged not guilty, it
does not mean that the person is proved innocent; it only means that there was not enough
evidence to reach the guilty verdict.
Example 8–4
A researcher claims that the average cost of men’s athletic shoes is less than $80.
He selects a random sample of 36 pairs of shoes from a catalog and finds the
following costs (in dollars). (The costs have been rounded to the nearest dollar.) Is there
enough evidence to support the researcher’s claim at a 0.10?
60
70
75
55
80
55
50
40
80
70
50
95
120
90
75
85
80
60
110
65
80
85
85
45
75
60
90
90
60
95
110
85
45
90
70
70
Solution
Step 1
State the hypotheses and identify the claim
H0: m $80
and
H1: m $80 (claim)
Step 2
Find the critical value. Since a 0.10 and the test is a left-tailed test, the
critical value is 1.28.
Step 3
Compute the test value. Since the exercise gives raw data, it is necessary to find
the mean and standard deviation of the data. Using the formulas in Chapter 3
or your calculator gives X 75.0 and s 19.2. Substitute in the formula
75 80
Xm
1.56
z
s
兹n
19.2
兹36
Make the decision. Since the test value, 1.56, falls in the critical region, the
decision is to reject the null hypothesis. See Figure 8–14.
Step 4
Figure 8–14
Critical and Test Values
for Example 8–4
–1.56 –1.28
Step 5
0
Summarize the results. There is enough evidence to support the claim that the
average cost of men’s athletic shoes is less than $80.
8–17
408
Chapter 8 Hypothesis Testing
Comment: In Example 8–4, the difference is said to be significant. However, when
the null hypothesis is rejected, there is always a chance of a type I error. In this case, the
probability of a type I error is at most 0.10, or 10%.
Example 8–5
The Medical Rehabilitation Education Foundation reports that the average cost of
rehabilitation for stroke victims is $24,672. To see if the average cost of rehabilitation
is different at a particular hospital, a researcher selects a random sample of 35 stroke
victims at the hospital and finds that the average cost of their rehabilitation is $25,226.
The standard deviation of the population is $3251. At a 0.01, can it be concluded
that the average cost of stroke rehabilitation at a particular hospital is different from
$24,672?
Source: Snapshot, USA TODAY.
Solution
Step 1
State the hypotheses and identify the claim.
H0: m $24,672
and
H1: m $24,672 (claim)
Step 2
Find the critical values. Since a 0.01 and the test is a two-tailed test, the
critical values are 2.58 and 2.58.
Step 3
Compute the test value.
z
Step 4
X m 25,226 24,672
1.01
s
兹n
3251
兹35
Make the decision. Do not reject the null hypothesis, since the test value falls
in the noncritical region, as shown in Figure 8–15.
Figure 8–15
Critical and Test Values
for Example 8–5
–2.58
Step 5
0
1.01
2.58
Summarize the results. There is not enough evidence to support the claim that
the average cost of rehabilitation at the particular hospital is different from
$24,672.
As with confidence intervals, the central limit theorem states that when the population standard deviation s is unknown, the sample standard deviation s can be used in
the formula as long as the sample size is 30 or more. The formula for the z test in this
case is
z
8–18
Xm
s
兹n
Section 8–3 z Test for a Mean
I. Claim is H0
Figure 8–16
Outcomes of a
Hypothesis-Testing
Situation
409
Reject H0
Do not reject H0
There is enough evidence
There is not enough evidence
to reject the claim.
to reject the claim.
II. Claim is H1
Reject H0
Do not reject H0
There is enough evidence
There is not enough evidence
to support the claim.
to support the claim.
When n is less than 30 and s is unknown, the t test must be used. The t test will be
explained in Section 8–4.
Students sometimes have difficulty summarizing the results of a hypothesis test.
Figure 8–16 shows the four possible outcomes and the summary statement for each
situation.
First, the claim can be either the null or alternative hypothesis, and one should identify which it is. Second, after the study is completed, the null hypothesis is either rejected
or not rejected. From these two facts, the decision can be identified in the appropriate
block of Figure 8–16.
For example, suppose a researcher claims that the mean weight of an adult animal
of a particular species is 42 pounds. In this case, the claim would be the null hypothesis, H0: m 42, since the researcher is asserting that the parameter is a specific value.
If the null hypothesis is rejected, the conclusion would be that there is enough evidence
to reject the claim that the mean weight of the adult animal is 42 pounds. See
Figure 8–17(a).
I. Claim is H0
Figure 8–17
Outcomes of a
Hypothesis-Testing
Situation for Two
Specific Cases
Reject H0
Do not reject H0
There is enough evidence
There is not enough evidence
to reject the claim.
to reject the claim.
(a) Decision when claim is H0 and H0 is rejected
II. Claim is H1
Reject H0
Do not reject H0
There is enough evidence
There is not enough evidence
to support the claim.
to support the claim.
(b) Decision when claim is H1 and H0 is not rejected
8–19
410
Chapter 8 Hypothesis Testing
On the other hand, suppose the researcher claims that the mean weight of the adult
animals is not 42 pounds. The claim would be the alternative hypothesis H1: m 42.
Furthermore, suppose that the null hypothesis is not rejected. The conclusion, then,
would be that there is not enough evidence to support the claim that the mean weight of
the adult animals is not 42 pounds. See Figure 8–17(b).
Again, remember that nothing is being proved true or false. The statistician is only
stating that there is or is not enough evidence to say that a claim is probably true or false.
As noted previously, the only way to prove something would be to use the entire population under study, and usually this cannot be done, especially when the population is large.
P-Value Method for Hypothesis Testing
Statisticians usually test hypotheses at the common a levels of 0.05 or 0.01 and sometimes at 0.10. Recall that the choice of the level depends on the seriousness of the
type I error. Besides listing an a value, many computer statistical packages give a
P-value for hypothesis tests.
The P-value (or probability value) is the probability of getting a sample statistic (such as
the mean) or a more extreme sample statistic in the direction of the alternative hypothesis
when the null hypothesis is true.
In other words, the P-value is the actual area under the standard normal distribution curve
(or other curve, depending on what statistical test is being used) representing the probability of a particular sample statistic or a more extreme sample statistic occurring if the
null hypothesis is true.
For example, suppose that a null hypothesis is H0: m 50 and the mean of a sample is X 52. If the computer printed a P-value of 0.0356 for a statistical test, then the
probability of getting a sample mean of 52 or greater is 0.0356 if the true population
mean is 50 (for the given sample size and standard deviation). The relationship
between the P-value and the a value can be explained in this manner. For P 0.0356,
the null hypothesis would be rejected at a 0.05 but not at a 0.01. See Figure 8–18.
When the hypothesis test is two-tailed, the area in one tail must be doubled. For
a two-tailed test, if a is 0.05 and the area in one tail is 0.0356, the P-value will be
2(0.0356) 0.0712. That is, the null hypothesis should not be rejected at a 0.05, since
0.0712 is greater than 0.05. In summary, then, if the P-value is less than a, reject the null
hypothesis. If the P-value is greater than a, do not reject the null hypothesis.
The P-values for the z test can be found by using Table E in Appendix C. First find
the area under the standard normal distribution curve corresponding to the z test value;
then subtract this area from 0.5000 to get the P-value for a right-tailed or a left-tailed test.
To get the P-value for a two-tailed test, double this area after subtracting. This procedure
is shown in step 3 of Examples 8–6 and 8–7.
Figure 8–18
Comparison of A
Values and P-Values
Area = 0.05
Area = 0.0356
Area = 0.01
50
8–20
52
Section 8–3 z Test for a Mean
411
The P-value method for testing hypotheses differs from the traditional method somewhat. The steps for the P-value method are summarized next.
Procedure Table
Solving Hypothesis-Testing Problems (P-Value Method)
Step 1
State the hypotheses and identify the claim.
Step 2
Compute the test value.
Step 3
Find the P-value.
Step 4
Make the decision.
Step 5
Summarize the results.
Examples 8–6 and 8–7 show how to use the P-value method to test hypotheses.
Example 8–6
A researcher wishes to test the claim that the average age of lifeguards in Ocean City
is greater than 24 years. She selects a sample of 36 guards and finds the mean of the
sample to be 24.7 years, with a standard deviation of 2 years. Is there evidence to
support the claim at a 0.05? Use the P-value method.
Solution
Step 1
State the hypotheses and identify the claim.
H0: m 24
Step 2
H1: m 24 (claim)
Compute the test value.
z
Step 3
and
24.7 24
2.10
2
兹36
Find the P-value. Using Table E in Appendix C, find the corresponding area
under the normal distribution for z 2.10. It is 0.4821. Subtract this value for
the area from 0.5000 to find the area in the right tail.
0.5000 0.4821 0.0179
Hence, the P-value is 0.0179.
Step 4
Make the decision. Since the P-value is less than 0.05, the decision is to reject
the null hypothesis. See Figure 8–19.
Figure 8–19
P-Value and A Value for
Example 8–6
Area = 0.05
Area = 0.0179
24
Step 5
24.7
Summarize the results. There is enough evidence to support the claim that the
average age of lifeguards in Ocean City is greater than 24 years.
8–21
412
Chapter 8 Hypothesis Testing
Note: Had the researcher chosen a 0.01, the null hypothesis would not have been
rejected, since the P-value (0.0179) is greater than 0.01.
Example 8–7
A researcher claims that the average wind speed in a certain city is 8 miles per hour.
A sample of 32 days has an average wind speed of 8.2 miles per hour. The standard
deviation of the sample is 0.6 mile per hour. At a 0.05, is there enough evidence to
reject the claim? Use the P-value method.
Solution
Step 1
State the hypotheses and identify the claim.
H0: m 8 (claim)
and
H1: m 8
Step 2
Compute the test value.
8.2 8
1.89
z
0.6
兹32
Step 3
Find the P-value. Using Table E, find the corresponding area for z 1.89. It
is 0.4706. Subtract the value from 0.5000.
0.5000 0.4706 0.0294
Since this is a two-tailed test, the area 0.0294 must be doubled to get the
P-value.
2 (0.0294) 0.0588
Step 4
Make the decision. The decision is not to reject the null hypothesis, since the
P-value is greater than 0.05. See Figure 8–20.
Figure 8–20
P-Values and A Values
for Example 8–7
Area = 0.0294
Area = 0.0294
Area = 0.025
Area = 0.025
8
Step 5
8.2
Summarize the results. There is not enough evidence to reject the claim that
the average wind speed is 8 miles per hour.
In Examples 8–6 and 8–7, the P-value and the a value were shown on a normal distribution curve to illustrate the relationship between the two values; however, it is not
necessary to draw the normal distribution curve to make the decision whether to reject
the null hypothesis. One can use the following rule:
Decision Rule When Using a P-Value
If P-value a, reject the null hypothesis.
If P-value a, do not reject the null hypothesis.
8–22
413
Section 8–3 z Test for a Mean
In Example 8–6, P-value 0.0179 and a 0.05. Since P-value a, the null hypothesis was rejected. In Example 8–7, P-value 0.0588 and a 0.05. Since P-value a,
the null hypothesis was not rejected.
The P-values given on calculators and computers are slightly different from those
found with Table E. This is due to the fact that z values and the values in Table E have
been rounded. Also, most calculators and computers give the exact P-value for two-tailed
tests, so it should not be doubled (as it should when the area found in Table E is used).
A clear distinction between the a value and the P-value should be made. The a value
is chosen by the researcher before the statistical test is conducted. The P-value is computed after the sample mean has been found.
There are two schools of thought on P-values. Some researchers do not choose an a
value but report the P-value and allow the reader to decide whether the null hypothesis
should be rejected.
In this case, the following guidelines can be used, but be advised that these guidelines are not written in stone, and some statisticians may have other opinions.
Guidelines for P-Values
If P-value 0.01, reject the null hypothesis. The difference is highly significant.
If P-value 0.01 but P-value 0.05, reject the null hypothesis. The difference is
significant.
If P-value 0.05 but P-value 0.10, consider the consequences of type I error before
rejecting the null hypothesis.
If P-value 0.10, do not reject the null hypothesis. The difference is not significant.
Others decide on the a value in advance and use the P-value to make the decision, as
shown in Examples 8–6 and 8–7. A note of caution is needed here: If a researcher selects
a 0.01 and the P-value is 0.03, the researcher may decide to change the a value from
0.01 to 0.05 so that the null hypothesis will be rejected. This, of course, should not be
done. If the a level is selected in advance, it should be used in making the decision.
One additional note on hypothesis testing is that the researcher should distinguish
between statistical significance and practical significance. When the null hypothesis is
rejected at a specific significance level, it can be concluded that the difference is probably not due to chance and thus is statistically significant. However, the results may not
have any practical significance. For example, suppose that a new fuel additive increases
the miles per gallon that a car can get by 14 mile for a sample of 1000 automobiles. The
results may be statistically significant at the 0.05 level, but it would hardly be worthwhile
to market the product for such a small increase. Hence, there is no practical significance
to the results. It is up to the researcher to use common sense when interpreting the results
of a statistical test.
Applying the Concepts 8–3
Car Thefts
You recently received a job with a company that manufactures an automobile antitheft device.
To conduct an advertising campaign for your product, you need to make a claim about the
number of automobile thefts per year. Since the population of various cities in the United
States varies, you decide to use rates per 10,000 people. (The rates are based on the number of
people living in the cities.) Your boss said that last year the theft rate per 10,000 people was
44 vehicles. You want to see if it has changed. The following are rates per 10,000 people for
36 randomly selected locations in the United States.
8–23
414
Chapter 8 Hypothesis Testing
55
39
51
15
70
58
42
69
55
53
25
56
125
23
26
56
62
33
62
94
66
91
115
75
134
73
41
20
17
20
73
24
67
78
36
16
Source: Based on information from the National Insurance Crime Bureau.
Using this information, answer these questions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
What are the hypotheses that you would use?
Is the sample considered small or large?
What assumption must be met before the hypothesis test can be conducted?
Which probability distribution would you use?
Would you select a one- or two-tailed test? Why?
What critical value(s) would you use?
Conduct a hypothesis test.
What is your decision?
What is your conclusion?
Write a brief statement summarizing your conclusion.
If you lived in a city whose population was about 50,000, how many automobile thefts
per year would you expect to occur?
See page 460 for the answers.
Exercises 8–3
For Exercises 1 through 13, perform each of the
following steps.
a. State the hypotheses and identify the claim.
b.
c.
d.
e.
Find the critical value(s).
Compute the test value.
Make the decision.
Summarize the results.
Use diagrams to show the critical region (or regions),
and use the traditional method of hypothesis testing
unless otherwise specified.
1. A survey claims that the average cost of a hotel room in
Atlanta is $69.21. To test the claim, a researcher selects a
sample of 30 hotel rooms and finds that the average cost is
$68.43. The standard deviation of the population is $3.72.
At a 0.05, is there enough evidence to reject the claim?
Source: USA TODAY.
2. It has been reported that the average credit card debt for
college seniors is $3262. The student senate at a large
university feels that their seniors have a debt much less
than this, so it conducts a study of 50 randomly selected
seniors and finds that the average debt is $2995 with a
sample standard deviation of $1100. With a 0.05, is
the student senate correct?
Source: USA TODAY.
8–24
3. A researcher estimates that the average revenue of
the largest businesses in the United States is greater
than $24 billion. A sample of 50 companies is selected,
and the revenues (in billions of dollars) are shown.
At a 0.05, is there enough evidence to support the
researcher’s claim?
178
122
91
44
35
61
30
29
41
31
24
25
24
22
56
28
16
38
30
16
25
23
21
46
28
16
36
19
15
18
17
20
20
20
19
15
19
15
14
17
17
32
27
15
25
19
19
15
22
20
Source: N.Y. Times Almanac.
4. Full-time Ph.D. students receive an average salary of
$12,837 according to the U.S. Department of Education.
The dean of graduate studies at a large state university
feels that Ph.D. students in his state earn more than this.
He surveys 44 randomly selected students and finds
their average salary is $14,445 with a standard deviation
of $1500. With a 0.05, is the dean correct?
Source: U.S. Department of Education/Chronicle of Higher Education.
5. A report in USA TODAY stated that the average age of
commercial jets in the United States is 14 years. An
Section 8–3 z Test for a Mean
executive of a large airline company selects a sample
of 36 planes and finds the average age of the planes is
11.8 years. The standard deviation of the sample is
2.7 years. At a 0.01, can it be concluded that the
average age of the planes in his company is less than
the national average?
Source: USA TODAY.
Source: The Old Farmers’ Almanac.
7. The average 1-year-old (both genders) is 29 inches
tall. A random sample of 30 one-year-olds in a large
day-care franchise resulted in the following heights. At
a 0.05, can it be concluded that the average height
differs from 29 inches?
25 32
35 25 30 26.5 26 25.5 29.5 32
30 28.5 30 32 28 31.5 29 29.5 30
34
29 32
27 28 33 28
27 32
29
29.5
12. The average salary for public school teachers for a
specific year was reported to be $39,385. A random
sample of 50 public school teachers in a particular state
had a mean of $41,680 and a standard deviation of
$5975. Is there sufficient evidence at the a 0.05 level
to conclude that the mean salary differs from $39,385?
Source: N.Y. Times Almanac.
13. To see if young men ages 8 through 17 years spend
more or less than the national average of $24.44 per
shopping trip to a local mall, the manager surveyed
33 young men and found the average amount spent per
visit was $22.97. The standard deviation of the sample
was $3.70. At a 0.02, can it be concluded that the
average amount spent at a local mall is not equal to the
national average of $24.44?
Source: USA TODAY.
14. What is meant by a P-value?
Source: www.healthepic.com.
8. At a certain university the mean income of parents
of the entering class is reported to be $91,600. The
president of another university feels that the parents’
income for her entering class is greater than $91,600.
She surveys 100 randomly selected families and finds
the mean income to be $96,321 with a standard
deviation of $9555. With a 0.05, is she correct?
Source: Chronicle of Higher Education.
9. Average undergraduate cost for tuition, fees, room, and
board for all institutions last year was $19,410. A random
sample of costs this year for 40 institutions of higher
learning indicated that the sample mean was $22,098,
and the sample standard deviation was $6050. At the
0.01 level of significance, is there sufficient evidence to
conclude that the cost of attendance has increased?
Source: N.Y. Times Almanac.
15. State whether the null hypothesis should be rejected on
the basis of the given P-value.
a.
b.
c.
d.
e.
P-value 0.258, a 0.05, one-tailed test
P-value 0.0684, a 0.10, two-tailed test
P-value 0.0153, a 0.01, one-tailed test
P-value 0.0232, a 0.05, two-tailed test
P-value 0.002, a 0.01, one-tailed test
16. A researcher claims that the yearly consumption of soft
drinks per person is 52 gallons. In a sample of 50
randomly selected people, the mean of the yearly
consumption was 56.3 gallons. The standard deviation
of the sample was 3.5 gallons. Find the P-value for the
test. On the basis of the P-value, is the researcher’s
claim valid?
Source: U.S. Department of Agriculture.
10. A real estate agent claims that the average price of a
home sold in Beaver County, Pennsylvania, is $60,000.
A random sample of 36 homes sold in the county is
selected, and the prices in dollars are shown. Is there
enough evidence to reject the agent’s claim at a 0.05?
54,000
121,500
13,000
7,500
92,000
28,000
85,000
11. The average U.S. wedding includes 125 guests. A
random sample of 35 weddings during the past year in
a particular county had a mean of 110 guests and a
standard deviation of 30. Is there sufficient evidence at
the 0.01 level of significance that the average number
of guests differs from the national average?
Source: www.theknot.com.
6. The average production of peanuts in the state of
Virginia is 3000 pounds per acre. A new plant food
has been developed and is tested on 60 individual
plots of land. The mean yield with the new plant food
is 3120 pounds of peanuts per acre with a standard
deviation of 578 pounds. At a 0.05, can one
conclude that the average production has increased?
9,500
29,000
6,000
42,000
95,000
15,000
76,000
284,000
415
99,000
184,750
188,400
32,900
38,000
53,500
25,225
Source: Pittsburgh Tribune-Review.
94,000
15,000
121,000
126,900
60,000
27,000
40,000
80,000
164,450
308,000
25,225
211,000
21,000
97,000
17. A study found that the average stopping distance of a
school bus traveling 50 miles per hour was 264 feet. A
group of automotive engineers decided to conduct a study
of its school buses and found that for 20 buses, the average stopping distance of buses traveling 50 miles per
hour was 262.3 feet. The standard deviation of the population was 3 feet. Test the claim that the average stopping
distance of the company’s buses is actually less than
264 feet. Find the P-value. On the basis of the P-value,
should the null hypothesis be rejected at a 0.01?
Assume that the variable is normally distributed.
Source: Snapshot, USA TODAY, March 12, 1992.
18. A store manager hypothesizes that the average
number of pages a person copies on the store’s copy
8–25
416
Chapter 8 Hypothesis Testing
machine is less than 40. A sample of 50 customers’
orders is selected. At a 0.01, is there enough
evidence to support the claim? Use the P-value
hypothesis-testing method.
2
5
21
21
3
37
9
80
21
3
2
29
1
85
15
5
2
9
49
17
2
8
24
61
27
3
1
51
36
17
5
2
72
8
113
58
6
2
43
4
32
49
70
42
36
82
9
122
61
1
19. A health researcher read that a 200-pound male can
burn an average of 546 calories per hour playing tennis.
Thirty-six males were randomly selected and tested. The
mean of the number of calories burned per hour was
544.8. Test the claim that the average number of calories
burned is actually less than 546, and find the P-value.
On the basis of the P-value, should the null hypothesis
be rejected at a 0.01? The standard deviation of the
sample is 3. Can it be concluded that the average number
of calories burned is less than originally thought?
20. A special cable has a breaking strength of 800 pounds.
The standard deviation of the population is 12 pounds.
A researcher selects a sample of 20 cables and finds
that the average breaking strength is 793 pounds.
Can one reject the claim that the breaking strength is
800 pounds? Find the P-value. Should the null
hypothesis be rejected at a 0.01? Assume that the
variable is normally distributed.
21. Several years ago the Department of Agriculture found
that the average size of farms in the United States
was 47.1 acres. A random sample of 50 farms was
selected, and the mean size of the farm was 43.2 acres.
The standard deviation of the sample was 8.6 acres. Test
the claim at a 0.05 that the average farm size is
smaller today, by using the P-value method. Should the
Department of Agriculture update its information?
22. Ten years ago, the average acreage of farms in a certain
geographic region was 65 acres. The standard deviation
of the population was 7 acres. A recent study consisting
of 22 farms showed that the average was 63.2 acres per
farm. Test the claim, at a 0.10, that the average has not
changed by finding the P-value for the test. Assume that
s has not changed and the variable is normally
distributed.
23. A car dealer recommends that transmissions be serviced
at 30,000 miles. To see whether her customers are
adhering to this recommendation, the dealer selects a
sample of 40 customers and finds that the average
mileage of the automobiles serviced is 30,456. The
standard deviation of the sample is 1684 miles. By
finding the P-value, determine whether the owners are
having their transmissions serviced at 30,000 miles.
Use a 0.10. Do you think the a value of 0.10 is an
appropriate significance level?
24. A motorist claims that the South Boro Police
issue an average of 60 speeding tickets per day. These
data show the number of speeding tickets issued each
day for a period of one month. Assume s is 13.42.
Is there enough evidence to reject the motorist’s claim
at a 0.05? Use the P-value method.
72
45
36
68
69
71
57
60
83
26
60
72
58
87
48
59
60
56
64
68
42
57
57
58
63
49
73
75
42
63
25. A manager states that in his factory, the average
number of days per year missed by the employees
due to illness is less than the national average of 10.
The following data show the number of days missed
by 40 employees last year. Is there sufficient evidence
to believe the manager’s statement at a 0.05?
(Use s to estimate s.) Use the P-value method.
0
6
12
3
3
5
4
3
9
6
0
7
6
3
7
4
7
1
0
8
12
2
5
10
5
15
3
2
3
11
8
2
2
4
1
Extending the Concepts
26. Suppose a statistician chose to test a hypothesis at
a 0.01. The critical value for a right-tailed test is
2.33. If the test value was 1.97, what would the
decision be? What would happen if, after seeing the test
value, she decided to choose a 0.05? What would the
decision be? Explain the contradiction, if there is one.
27. The president of a company states that the average
hourly wage of her employees is $8.65. A sample of
50 employees has the distribution shown. At a 0.05,
8–26
is the president’s statement believable? (Use s to
approximate s.)
Class
Frequency
8.35–8.43
8.44–8.52
8.53–8.61
8.62–8.70
8.71–8.79
8.80–8.88
2
6
12
18
10
2
1
4
3
5
9
Section 8–3 z Test for a Mean
417
Technology Step by Step
MINITAB
Hypothesis Test for the Mean and the z Distribution
Step by Step
MINITAB can be used to calculate the test statistic and its P-value. The P-value approach does
not require a critical value from the table. If the P-value is smaller than a, the null hypothesis
is rejected. For Example 8–4, test the claim that the mean shoe cost is less than $80.
1. Enter the data into a column of MINITAB. Do not try to type in the dollar signs! Name the
column ShoeCost.
2. If sigma is known, skip to step 3; otherwise estimate sigma from the sample standard
deviation s.
Calculate the Standard Deviation in the Sample
a) Select Calc >Column Statistics.
b) Check the button for Standard deviation.
c) Select ShoeCost for the Input variable.
d) Type s in the text box for Store the result in:.
e) Click [OK].
Calculate the Test Statistic and P-Value
3. Select Stat >Basic Statistics>1 Sample Z, then select ShoeCost in the Variable text
box.
4. Click in the text box and enter the value of sigma or type s, the sample standard deviation.
5. Click in the text box for Test mean, and enter the hypothesized value of 80.
6. Click on [Options].
a) Change the Confidence level to 90.
b) Change the Alternative to less than. This setting is crucial for calculating the P-value.
7. Click [OK] twice.
One-Sample Z: ShoeCost
Test of mu = 80 vs < 80
The assumed sigma 19.161
Variable
ShoeCost
N
36
Mean
75.0000
StDev
19.1610
SE Mean
3.1935
90%
Upper
Bound
79.0926
Z
-1.57
P
0.059
Since the P-value of 0.059 is less than a, reject the null hypothesis. There is enough evidence
in the sample to conclude the mean cost is less than $80.
8–27
418
Chapter 8 Hypothesis Testing
TI-83 Plus or
TI-84 Plus
Step by Step
Hypothesis Test for the Mean and the z Distribution (Data)
1. Enter the data values into L1.
2. Press STAT and move the cursor to TESTS.
3. Press l for ZTest.
4. Move the cursor to Data and press ENTER.
5. Type in the appropriate values.
6. Move the cursor to the appropriate alternative hypothesis and press ENTER.
7. Move the cursor to Calculate and press ENTER.
Example TI8–1
This relates to Example 8–4 from the text. At the 10% significance level, test the claim
that m 80 given the data value.
60
120
75
70
90
60
75
75
90
55
85
90
80
80
60
55
60
95
50
110
110
40
65
85
80
80
45
70
85
90
50
85
70
95
45
70
The population standard deviation s is unknown. Since the sample size n 36 30, one can
use the sample standard deviation s as an approximation for s. After the data values are
entered in L1 (step 1), press STAT, move the cursor to CALC, press 1 for 1-Var Stats, then
press ENTER. The sample standard deviation of 19.16097224 will be one of the statistics
listed. Then continue with step 2. At step 5 on the line for s press VARS for variables, press 5
for Statistics, press 3 for Sx.
The test statistic is z 1.565682556, and the P-value is 0.0587114841.
Hypothesis Test for the Mean and the z Distribution (Statistics)
1. Press STAT and move the cursor to TESTS.
2. Press 1 for ZTest.
3. Move the cursor to Stats and press ENTER.
4. Type in the appropriate values.
5. Move the cursor to the appropriate alternative hypothesis and press ENTER.
6. Move the cursor to Calculate and press ENTER.
Example TI8–2
This relates to Example 8–3 from the text. At the 5% significance level, test the claim that
m 42,000 given s 5230, X 43,260, and n 30.
The test statistic is z 1.319561037, and the P-value is 0.0934908728.
Excel
Hypothesis Test for the Mean and the z Distribution
Step by Step
Excel does not have a procedure to conduct a hypothesis test for the mean. However, you may
conduct the test of the mean using the MegaStat Add-in available on your CD and Online Learning
Center. If you have not installed this add-in, do so by following the instructions on page 24.
This test uses the P-value method. Therefore, it is not necessary to enter a significance level.
1. Enter the data from Example 8–4 into column A of a new worksheet.
2. Select MegaStat >Hypothesis Tests>Mean vs. Hypothesized Value.
3. Select data input and enter the range of data A1:A36.
4. Type 80 for the Hypothesized mean and select the “less than” Alternative.
5. Select z test and click [OK].
8–28